OFFSET
1,2
COMMENTS
This is the case m = 3 of the general recurrence a(1) = 1, a(2) = 2*m+3, a(n+2) = (2*m+3)*a(n+1) + (n+1)*(n+3)*a(n), which arises when accelerating the convergence of a certain series for the constant log(2). For remarks on the general case see A142988 (m=0). For other cases see A142989 (m=1) and A142990 (m=2).
LINKS
Seiichi Manyama, Table of n, a(n) for n = 1..445
FORMULA
a(n) = (n+2)!*p(n+2)*sum {k = 1..n} (-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), where p(n) = (2*n^3-3*n^2+7*n-3)/15. Recurrence: a(1) = 1, a(2) = 9, a(n+2) = 9*a(n+1)+(n+1)*(n+3)*a(n). The sequence b(n) := 1/2*(n+2)!*p(n+2) satisfies the same recurrence with b(1) = 9, b(2) = 84. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(9+1*3/(9+2*4/(9+3*5/(9+...+(n-1)*(n+1)/9)))), for n >=2. Lim n -> infinity a(n)/b(n) = 1/(9+1*3/(9+2*4/(9+3*5/(9+...+(n-1)*(n+1)/(9+...))))) = 2*sum {k = 1..inf} (-1)^(k+1)/ (k*(k+1)*(k+2)*p(k+1)*p(k+2)) = 167/6 - 40*log(2).
MAPLE
p := n -> (2*n^3-3*n^2+7*n-3)/15: a := n -> (n+2)!*p(n+2)*sum ((-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), k = 1..n): seq(a(n), n = 1..20);
MATHEMATICA
RecurrenceTable[{a[1]==1, a[2]==9, a[n+2]==9a[n+1]+(n+1)(n+3)a[n]}, a, {n, 20}] (* Harvey P. Dale, Jul 18 2020 *)
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Peter Bala, Jul 17 2008
STATUS
approved